Prove that if a function is nonnegative on an interval, so is its definite integral

[Meow12]

If [imath]f(x)\ge 0[/imath] on [imath][a,\ b][/imath], then prove that [imath]\int_a^b f(x)dx\ge 0[/imath].

My attempt:

Case 1: [imath]f(x)=0[/imath] on [imath][a,\ b][/imath].

[imath]G(x)=0\forall x\in [a,\ b][/imath] is an antiderivative of [imath]f(x)[/imath] on [imath][a,\ b][/imath].

[imath]\int_a^b f(x)dx=G(a)-G(b)=0-0=0[/imath]

Case 2: [imath]f'(x)>0[/imath] on [imath][a,\ b][/imath].

Let [imath]G(x)[/imath] be an antiderivative of [imath]f(x)[/imath] on [imath][a,\ b][/imath].

[imath]G'(x)=f(x)>0[/imath] on [imath][a,\ b][/imath].

So, [imath]G[/imath] is increasing on [imath][a,\ b][/imath].

Since [imath]a<b[/imath], [imath]G(a)<G(b)[/imath].

[imath]\implies\int_a^b f(x)dx=G(b)-G(a)>0[/imath]
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Thus, in both cases, [imath]\int_a^b f(x)dx\ge 0[/imath].

Does this look right?

 

[blamocur]

Looks good to me. But I'd try transforming Case 2 by replacing [imath]>[/imath] with [imath]\geq[/imath]. This way you don't need to consider Case 1.

 

[Meow12]

Looks good to me. But I'd try transforming Case 2 by replacing [imath]>[/imath] with [imath]\geq[/imath]. This way you don't need to consider Case 1.

Thanks!

The criterion for an increasing function is [imath]G'(x)>0[/imath], not [imath]G'(x)\ge 0[/imath]; that's why I had to consider two cases.

 

[Steven G]

I have a few problems with your case 1. Luckily most of your mistakes were actually ok. For example 0 -0 =0 - 0
\(\displaystyle \int_{a}^{b}{f(x)dx}= G(b)-G(a) = 0-0=0\). You wrote G(a)-G(b)

You said that an antiderivative of 0 is 0 which is true. Then you went on assuming that 0 is the only antiderivative of 0. This is NOT true! \(\displaystyle \int{0 dx} = C\), where C is any constant!

Case 2: Consider the line f(x)=2x. Clearly its derivative f'(x)=2x > 0 for all x.
Now consider the case where we restrict the interval to [-8, -2]. Note that f(x) <0 for all x in [-8,-2] while f'(x)>0 for all x in (-8,-2). This shows that what you are saying is not true at all--sorry.


Why does the two cases cover everything? Looking at case 1 is good!
However, f'(x)>0 is always true whenever f(x)>0 is NOT true at all.
Consider f(x) = x^2 + 1. This function is ALWAYS positive. Now note, that its derivative, f'(x)=2x, is NEGATIVE if x<0.

You need to see the following clearly. Knowing the sign of f(x) (that is if f(x)>0 or f(x)<0) does NOT tell you any information about whether f'(x)>0 or f'(x)<0. Similarly, knowing the sign of f'(x) tell you nothing about the sign of f(x). Please post back with graphs showing this. You really have to understand this!

 

[Steven G]

Did you mean to write Case 2: f′(x)>0[a, b] or Case 2: f(x)>0[a, b]??
If writing f'(x) was a typo for f(x), then your proof is fine. That little dash makes a major difference!

 

[Meow12]

Did you mean to write Case 2: f′(x)>0[a, b] or Case 2: f(x)>0[a, b]??
If writing f'(x) was a typo for f(x), then your proof is fine. That little dash makes a major difference!

Yeah, that was a typo. Sorry.

 

[Meow12]

I have a few problems with your case 1. Luckily most of your mistakes were actually ok. For example 0 -0 =0 - 0
\(\displaystyle \int_{a}^{b}{f(x)dx}= G(b)-G(a) = 0-0=0\). You wrote G(a)-G(b)

You said that an antiderivative of 0 is 0 which is true. Then you went on assuming that 0 is the only antiderivative of 0. This is NOT true! \(\displaystyle \int{0 dx} = C\), where C is any constant!

In evaluating a definite integral, isn't it sufficient to find just one antiderivative?

For example, [imath]\int_0^1 3x^2 dx=[x^3+2]_0^1=3-2=1[/imath] where [imath]x^3+2[/imath] is one antiderivative of [imath]3x^2[/imath].

 

[Steven G]

In evaluating a definite integral, isn't it sufficient to find just one antiderivative?

For example, [imath]\int_0^1 3x^2 dx=[x^3+2]_0^1=3-2=1[/imath] where [imath]x^3+2[/imath] is one antiderivative of [imath]3x^2[/imath].

That is a very good point! Good work!

 

[khansaheb]

If f(x)≥0f(x)\ge 0f(x)≥0 on [a, b][a,\ b][a, b], then prove that ∫abf(x)dx≥0\int_a^b f(x)dx\ge 0∫abf(x)dx≥0.

Would it be valid if a < 0 ?